Model 4 Train Vs Pole/Signal Post/Man Practice Questions Answers Test with Solutions & More Shortcuts
trains PRACTICE TEST [6 - EXERCISES]
Model 1 Train Vs Train in same direction
Model 2 Train Vs Train in opposite direction
Model 3 Train Vs Bridge/Platform
Model 4 Train Vs Pole/Signal Post/Man
Model 5 Train Vs Both platform and a man/a pole
Model 6 Change in speed Vs Change with time travel
Question : 1 [SSC CPO S.I.2003]
A man observed that a train 120 m long crossed him in 9 seconds. The speed (in km/hr) of the train was
a) 45
b) 48
c) 55
d) 42
Answer »Answer: (b)
Using Rule 1,
Speed of train
= $120/9 × 18/5$ = 48 kmph
Question : 2 [SSC CGL Tier-I 2010]
A train, 300m long, passed a man, walking along the line in the same direction at the rate of 3 km/hr in 33 seconds. The speed of the train is
a) 32 km/h
b) 32$8/11$ km/h
c) 35$8/11$ km/h
d) 30 km/h
Answer »Answer: (c)
Using Rule 5,
If the speed of the train be x kmph,
then relative speed = (x - 3) kmph.
or (x - 3) × $5/18$ m/sec
$300/{(x - 3) × 5/18 = 33$
5400 = 33 × 5 (x - 3)
360 = 11 (x - 3)
11x - 33 = 360
$x = 393/11 = 35{8}/11$ kmph
Question : 3 [SSC CGL Tier-II 2015]
If a man walks at the rate of 5 km/hour, he misses a train by 7 minutes. However if he walks at the rate of 6 km/hour, he reaches the station 5 minutes before the arrival of the train. The distance covered by him to reach the station is
a) 7 km
b) 6.25 km
c) 4 km
d) 6 km
Answer »Answer: (d)
Let the required distance be x km.
Difference of time = 7 + 5 = 12
minutes = $1/5$ hour
Time = $\text"Distance"/ \text"Speed"$
According to the question,
$x/5 - x/6 = 1/5$
${6x - 5x}/30 = 1/5$
$x/30 = 1/5$ ⇒ $x = 30/5$ = 6 km.
Question : 4 [SSC CGL Tier-I 2010]
A train, 240 m long crosses a man walking along the line in opposite direction at the rate of 3 kmph in 10 seconds. The speed of the train is
a) 75 kmph
b) 83.4 kmph
c) 86.4 kmph
d) 63 kmph
Answer »Answer: (b)
Using Rule 6,
Let 'a' metre long train is running with the speed 'x' m/s. A man is running in the opposite direction of train with the speed of 'y' m/s. Then, time taken by the train to cross the man = $(a/{(x + y)})$seconds.
If the speed of train be x kmph then,
Its relative speed = (x + 3) kmph
Time = $\text"Length of the train"/ \text"Relative speed"$
$10/3600 = {240/1000}/{(x + 3)} = 240/{1000(x + 3)}$
x + 3 = 86.4 ⇒ x = 83.4 kmph
Question : 5 [SSC CGL Prelim 2002]
A 75 metre long train is moving at 20 kmph. It will cross a man standing on the platform in
a) 14 seconds
b) 13.5 seconds
c) 15.5 seconds
d) 12 seconds
Answer »Answer: (b)
Using Rule 1,
Speed of train (in m/s)
= 20 × $5/18 = 50/9$ m/sec
Required time = $75/50 × 9$
= 13.5 seconds
IMPORTANT quantitative aptitude EXERCISES
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Top 399+ Trains Based Time and Distance MCQs For BANK SSC »
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New 500+ Aptitude Problems on Train For All BANK SSC Exam »
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Top 499+ Aptitude MCQs on Trains Crossing Bridge/Platform »
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New 500+ Aptitude MCQs on Train Crossing Pole/Post or Man »
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Top 499+ Trains Problems Using Time and Distance For SSC »
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Top 489+ Time And Distance MCQ Problems on Train For BANK »
Model 4 Train Vs Pole/Signal Post/Man Shortcuts »
Click to Read...Model 4 Train Vs Pole/Signal Post/Man Online Quiz
Click to Start..trains Shortcuts and Techniques with Examples
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Model 1 Train Vs Train in same direction
Defination & Shortcuts … -
Model 2 Train Vs Train in opposite direction
Defination & Shortcuts … -
Model 3 Train Vs Bridge/Platform
Defination & Shortcuts … -
Model 4 Train Vs Pole/Signal Post/Man
Defination & Shortcuts … -
Model 5 Train Vs Both platform and a man/a pole
Defination & Shortcuts … -
Model 6 Change in speed Vs Change with time travel
Defination & Shortcuts …
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